Let \(\alpha ,\beta \in  have modulus less than \) 1/T\(.Then by \cite {YoungSL} we may write{\begin{@align*}{1}{-1}I_T(\alpha ,\beta ) := \int _T^{2T} \zeta (\tfrac{1}{2} + \alpha + i t)&\zeta (\tfrac{1}{2} + \beta + i t) |A(\tfrac{1}{2} + i t)|^2 dt \\= \sum _{h,k\le T^\theta } \frac{a_h \overline{a_k}}{[h,k]} \int _\mathbb {R}\Bigg ( \frac{(h,k)^{\alpha + \beta }}{h^\alpha k^\beta } \zeta (1 + \alpha + \beta ) + &\left(\frac{t}{2\pi }\right)^{-\alpha - \beta } \frac{(h,k)^{-\alpha - \beta }}{h^{-\alpha } k^{-\beta }} \zeta (1 - \alpha - \beta ) \Bigg ) \phi (t/T) dt + O(T^{1-\delta })\end{@align*}}for some \) >0\(, which is holomorphic in \) ,\( sufficiently small.We may express the main term as a multiple contour integral around \)\( and \)\(: by lemma 2.5.1 of \cite {CFKRS} and a shift of contours we find{\begin{@align*}{1}{-1}I_T(\alpha ,\beta ) =& -\frac{1}{(2\pi i)^2} \oint \limits _{|z_2 - \beta | = 9/\log T}\oint \limits _{|z_1 - \alpha | = 3/\log T} F(z_1, -z_2) \frac{\zeta (1+z_1-z_2) (z_1-z_2)^2}{(z_1 - \alpha )(z_1 + \beta )(z_2 - \alpha )(z_2 + \beta )} \\&\qquad \qquad \times \left(\int _\mathbb {R}\left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2 - \beta - \alpha }{2}} \phi (t/T) dt \right) dz_1 dz_2 + O(T^{1-\delta }) \\&= -\frac{1}{(2\pi i)^2} \oint \limits _{|z_2| = 9/\log T} \oint \limits _{|z_1| = 3/\log T} F(z_1, -z_2) \frac{\zeta (1+z_1-z_2) (z_1-z_2)^2}{(z_1 - \alpha )(z_1 + \beta )(z_2 - \alpha )(z_2 + \beta )} \\&\qquad \qquad \times \left(\int _\mathbb {R}\left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2 - \beta - \alpha }{2}} \phi (t/T) dt \right) dz_1 dz_2 + O(T^{1-\delta }).\end{@align*}}Note we do not cross any poles when shifting contours since \) ||,|| < 1/T\(.Now since \) IT(,)\( is holomorphic with respect to small \)\( and \)\(, as in \cite {YoungSL} the derivatives of \) IT(, )\( with respect to \)\( and \)\( can be obtained via Cauchy^{\prime }s theorem as contour integrals along circles of radii \) 1/T\(.Since the error term holds uniformly on these contours, we conclude{\begin{@align*}{1}{-1}\widetilde{I}_T(&\alpha ,\beta ) := \int _T^{2T} \zeta ^{\prime }(\tfrac{1}{2} + \alpha + i t)\zeta ^{\prime }(\tfrac{1}{2} + \beta + i t) |A(\tfrac{1}{2} + i t)|^2 dt\\= \frac{d}{d\alpha } \frac{d}{d\beta } \Bigg [ -\frac{1}{(2\pi i)^2} &\oint \limits _{|z_2| = 9/\log T}\oint \limits _{|z_1| = 3/\log T} F(z_1, -z_2) \frac{\zeta (1+z_1-z_2) (z_1-z_2)^2}{(z_1-\alpha )(z_1 + \beta )(z_2 - \alpha )(z_2 + \beta )} \\&\times \left(\int _\mathbb {R}\left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2 - \beta - \alpha }{2}} \phi (t/T) dt \right) dz_1 dz_2 \Bigg ] + O(T^{1-\delta }).\end{@align*}}To compute \) I(1)(T)\(, we evaluate these derivatives and then set \) = = 0\(, obtaining{\begin{@align*}{1}{-1}\widetilde{I}^{(1)}(T) =& \frac{1}{(2\pi i)^2} \oint \limits _{|z_2| = 9/\log T} \oint \limits _{|z_1| = 3/\log T} F(z_1, -z_2) \zeta (1+z_1-z_2) (z_1-z_2)^2 \int _\mathbb {R}\Bigg [\left(z_1 + z_2 + \frac{z_1 z_2}{2} \log \frac{t}{2\pi } \right)\\& \times \left(z_1 + z_2 - \frac{z_1 z_2}{2} \log \frac{t}{2\pi }\right) \left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2}{2}} \phi (t/T) dt \Bigg ] \frac{dz_1}{z_1^4}\frac{dz_2}{z_2^4} + O(T^{1-\delta }).\end{@align*}}Finally, since \) |zj| = 3j/T\( and \)  supp [3/4,9/4]\(, notice that\)\(\zeta (1+z_1 - z_2) \ll \log T,\quad (z_1-z_2)^2 \ll (\log T)^{-2},\)\(and\)\(\int _\mathbb {R}\left(z_1 + z_2 + \frac{z_1 z_2}{2} \log \frac{t}{2\pi } \right) \left(z_1 + z_2 - \frac{z_1 z_2}{2} \log \frac{t}{2\pi }\right) \left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2}{2}} \phi (t/T) dt \ll T(\log T)^{-2},\)\(so the claim now follows.\) The case for twisted moments of \(Z\)  is similar.
The main difference is that applying lemma 2.5.1 of [1]} gives up to a power savings the simpler formula
\(-\frac{1}{(2\pi i)^2} &\oint \limits _{|z_2| = 9/\log T}\oint \limits _{|z_1| = 3/\log T} F(z_1, -z_2) \frac{\zeta (1+z_1-z_2) (z_1-z_2)^2}{(z_1 - \alpha )(z_1 + \beta )(z_2 - \alpha )(z_2 + \beta )} \\&\qquad \qquad \times \left(\int _\mathbb {R}\left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2}{2}} \phi (t/T) dt \right) dz_1 dz_2.\) 

Then differentiating with respect to \(\alpha \)  and \(\beta \)  and setting the shifts to zero we obtain
\(\frac{1}{(2\pi i)^2} &\oint \limits _{|z_2| = 9/\log T} \oint \limits _{|z_1| = 3/\log T} F(z_1, -z_2) \zeta (1+z_1-z_2) (z_1^2-z_2^2)^2 \left(\int _\mathbb {R}\left(\frac{t}{2\pi }\right)^{\tfrac{z_1 - z_2}{2}} \phi (t/T) dt \right) \frac{dz_1}{z_1^4}\frac{dz_2}{z_2^4},\) 

which satisfies the same bound.
